How many integers $n$ satisfy the condition $100 < n < 200$ and the condition $n$ has the same remainder whether it is divided by $6$ or by $8$?
Since $n$ has the same remainder whether divided by 6 or by 8, we can write that $n = 6a + r = 8b + r$, where $0\leq r \leq 5$.  This implies that $3a = 4b$, and so $a$ is a multiple of 4 and we can write $a = 4k$ for some integer $k$.  Since $100<n<200$, we see that $95<6a<200$, or $\frac{95}{24} < k <\frac{200}{24}$.  Since $k$ is an integer, $4\leq k \leq 8$.  If $k = 4$, then we must have $r = 5$.  Otherwise, $r = 0,1,2,3,4,5$ are all allowable.  Thus we have a total of $\boxed{25}$ possible values for $n$.